Arrange the following ligands in decreasing order of field strength in spectrochemical series.

(A) $Br^-$
(B) $H_2O$
(C) $OH^-$
(D) $CO$

Choose the correct answer from the options given below:

(D), (B), (C), (A)

The correct answer is Option (1) → (D), (B), (C), (A)

Concept: Spectrochemical Series

Ligands are arranged based on the crystal field splitting ($\Delta$) they produce:

• Strong field ligands $\rightarrow$ large $\Delta \rightarrow$ low spin
• Weak field ligands $\rightarrow$ small $\Delta \rightarrow$ high spin

Relevant order:

$\mathbf{CO > OH^- > H_2O > Br^-} \text{}$

Option-wise Analysis

Option (A) $\text{Br}^-$: Bromide is a weak field ligand. It is large and highly polarizable, giving small crystal field splitting ($\Delta$). It generally leads to high-spin complexes.

Option (B) $\text{H}_2\text{O}$: Water is a neutral ligand with moderate donor ability. It produces greater splitting than halides but is still considered a medium field ligand in the spectrochemical series.

Option (C) $\text{OH}^-$: Hydroxide is a stronger donor than water because of its negative charge. It causes more repulsion with d-electrons and gives a stronger field than $\text{H}_2\text{O}$, but still weaker than strong $\pi$-acceptors.

Option (D) $\text{CO}$: Carbon monoxide is a strong field ligand. It is a good $\sigma$-donor and strong $\pi$-acceptor, causing large crystal field splitting and often forming low-spin complexes.